A particle starts from rest and traverses a distance 2x with uniform acceleration, then moves uniformly over a further distance 4x and finally comes to rest after moving a further distance 6x under uniform retardation. Assuming entire motion to be rectilinear motion, the ratio of average speed over the journey to the maximum speed on its way is

Question

A particle starts from rest and traverses a distance 2x with uniform acceleration, then moves uniformly over a further distance 4x and finally comes to rest after moving a further distance 6x under uniform retardation. Assuming entire motion to be rectilinear motion, the ratio of average speed over the journey to the maximum speed on its way is (A) (4/5) (B) (3/5) (C) (2/5) (D) (1/5)

Tardigrade Admin · Accepted Answer

Let t1 , t2 and t3 be the time taken by the particles to cover the distance 2x , 4x and 6x respectively. Let v be the velocity of the particle at B i.e, maximum velocity. The particle moves with uniform acceleration from A to B . For motion from A to B . Average velocity =(0 + v/2)=(v/2) Time taken, t1=(2 x/v/2) Particle moves with uniform retardation from C TO D , Va v g=(0 + v/2)=v/2 Time taken , t3=(6 x/(. v/2 .)) Total time = t1+t2+t3 t=(2 x/v/2)+(4 x/v)+(6 x/v/2)=(20 x/v) Average Velocity over= (6 x + 2 x + 4 x/20 x/v)=(3/5)v Required Ratio = (Average Speed/v)=(3/5)

$\frac{4}{5}$

$\frac{3}{5}$

$\frac{2}{5}$

$\frac{1}{5}$

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$\frac{4}{5}$

$\frac{3}{5}$

$\frac{2}{5}$

$\frac{1}{5}$